SOLVING ALGEBRA PROBLEMS - LINKS TO THEIR STUDY, SOLVED EXAMPLES, EXERCISES ON LOGARITHMS

Your Ad Here

Please study

Logarithm Formulas before Solving Algebra Problems

if you have not already done so.

Having the knowledge of the Formulae
is a prerequisite here.





Solved Example 1 of Solving Algebra Problems :

Find the value of
(i) log343 49 (ii) log0.01 (0.0001) (iii) log(2√3) (1728)

Solution to Example 1 of Solving Algebra Problems :

(i) Let A = log343 49
We know 49 = 7 x 7 = 72; 343 = 49 x 7 = 7 x 7 x 7 = 73
∴ A = log343 49 = log(73) (72)
We know, in log of a power to the base of another power (See Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ A = (2⁄3) log7 7
But log of any number to the same Base is one (see Formula 4).
∴ A = (2⁄3) (1) = 2⁄3. Ans.

(ii) Let A = log0.01 (0.0001)
We know 0.0001 = 0.01 x 0.01 = (0.01)2;
∴ A = log(0.01) (0.0001) = log(0.01) {(0.01)2}
We know, in log of a power (see Formula 7), the exponent multiplies the log.
∴ A = 2 log(0.01) (0.01)
But log of any number to the same Base is one (see Formula 4).
∴ A = 2 (1) = 2. Ans.

(iii) Let A = log(2√3) (1728)
We know 2√3 = 2 x (3)(1⁄2) = (22)(1⁄2) x (3)(1⁄2)
= 4(1⁄2) x (3)(1⁄2) = (4 x 3)(1⁄2) = 12(1⁄2)
1728 = 12 x 144 = 12 x 12 x 12 = 123
See how the base (2√3) and 1728 are made as powers of the same number 12.
∴ A = log(2√3) (1728) = log{12(1⁄2)} (123)
We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ A = {3⁄(1⁄2)} log12 12
But log of any number to the same Base is one (see Formula 4).
∴ A = (3 x 2) (1) = 6.
Thus log(2√3) (1728) = 6. Ans.





Get The Best Grades With the Least Amount of Effort :
Solving Algebra Problems

Here is a collection of proven tips,
tools and techniques to turn you into
a super-achiever - even if you've never
thought of yourself as a "gifted" student.

The secrets will help you absorb, digest
and remember large chunks of information
quickly and easily so you get the best grades
with the least amount of effort.

If you apply what you read from the above
collection, you can achieve best grades without
giving up your fun, such as TV, surfing the net,
playing video games or going out with friends!

Know more about the Speed Study System.

Great deals on School & Homeschool Curriculum Books and Software

Solved Example 2 of Solving Algebra Problems :

Solve the following Problem on Solving Algebra Problems :

If (2.3)x = (0.23)y = 1000, show that 1⁄x - 1⁄y = 1⁄3

Solution to Example 2 of Solving Algebra Problems :

By data (2.3)x = 1000
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
x = log(2.3) 1000
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄x
Reciprocal of R.H.S. = 1⁄(log(2.3) 1000)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log1000 (2.3)
∴ 1⁄x = log1000 (2.3)

By data (0.23)y = 1000
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
y = log(0.23) 1000
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄y
Reciprocal of R.H.S. = 1⁄(log(0.23) 1000)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log1000 (0.23)
∴ 1⁄y = log1000 (0.23)

To prove 1⁄x - 1⁄y = 1⁄3
L.H.S. = 1⁄x - 1⁄y = log1000 (2.3) - log1000 (0.23)
We know, log of a quotient can be written as the difference of the log of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ L.H.S. = log1000 {(2.3)⁄(0.23)} = log1000 (10)
= log103 (101)
We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.
∴ L.H.S. = (1⁄3) log10 10
We know log of a quantity to the same base is 1. (See Formula 4)
∴ L.H.S. = (1⁄3) (1) = 1⁄3 = R.H.S. (Proved.)

Great Deals on School & Homeschool Curriculum Books

Logarithm Tables : Characteristic and Mantissa

To find the log value of a number using Tables
we need the knowledge of Characteristic and Mantissa
which is covered in Logarithm Tables.

Research-based personalized
Math Help tutoring program :
Solving Algebra Problems

Here is a resource for Solid Foundation in
Math Fundamentals from Middle thru High School.
You can check your self by the

FREE TRIAL.

Are you spending lot of money for math tutors to your
child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and
more personalized Math Help to identify and fix
the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend
and that too With a minuscule fraction of the amount
you spent on tutors with unconditional 100% money
back Guarantee, if you are not satisfied.

SUBSCRIBE, TEST, IF NOT SATISFIED, RETURN FOR FULL REFUND

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program
that identifies exactly where your child needs help and
then creates a personal instruction plan tailored to your
child’s specific needs.

If your child can use a computer and access
the Internet, he or she can use the program.
And your child can access the program anytime
from any computer with Internet access.

Unique program to help improve math skills quickly and painlessly.

There is an exclusive, Parent Information Page provides YOU
with detailed reports of your child’s progress so you can
monitor your child’s success and give them encouragement.
These Reports include

  • Time spent using the program
  • Assessment results
  • Personalized remediation curriculum designed for your child
  • Details the areas of weakness where your child needs additional help
  • Provides the REASONS WHY your child missed a concept
  • List of modules accessed and amount of time spent in each module
  • Quiz results
  • Creates reports that can be printed and used to discuss issues with your child’s teachers
These reports are created and stored in a secure section
of the program, available exclusively to you, the parent.
The section is accessed by a password that YOU create and use.
No unauthorized users can access this information.

Personalized remediation curriculum designed for your child

Thus The features of this excellent Tutoring program are

  • Using detailed testing techniques
  • Identifing exactly where a student needs help.
  • Its unique, smart system pinpointing precise problem areas -
  • slowly and methodically guiding the student
  • raising to the necessary levels to fix the problem.

Not a “one-size-fits-all” approach!

Its research-based results have proven that
it really works for all students! in improving
math skills and a TWO LETTER GRADE INCREASE in
math test scores!,if they invest time in using
the program.

Proven for More than 10,000 U.S. public school
students who increased their math scores.

Proven methodology!



Exercise : Solving Algebra Problems

Solve the following problems on Solving Algebra Problems

  1. If (log2 a)⁄4 = (log2 b)⁄6 = (log2 c)⁄(3p) and a3.b2.c = 1,
    find the value of p
  2. If 2 logx a + logax a + 3 loga2x a = 0,
    find the value of x.

For Answers see at the bottom of the page.





Progressive Learning of Math : Solving Algebra Problems

Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.

These also provide teachers and parents with lessons
on how to work with the child on the concepts.

The series is low to reasonably priced and include

Elementary Math curriculum

and

Algebra Curriculum.


Answers to Exercise : Solving Algebra Problems

Answers to Problems on Solving Algebra Problems :
(1) -8 (2) a-1⁄2, a-4⁄3